Samacheer Class 9 Maths - Algebra

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# (i)

Solution

The exponent is negative.

Hence, it is not a polynomial.

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# (ii)

Solution

All exponents are non-negative integers.

Hence, it is a polynomial.

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# (iii)

Solution

Contains negative exponent.

Hence, it is not a polynomial.

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# (iv)

Solution

Variable occurs in denominator.

Hence, it is not a polynomial.

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# (v)

Solution

All exponents are non-negative integers.

Irrational coefficients are allowed.

Hence, it is a polynomial.

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# (vi)

Solution

Exponent is fractional.

Hence, it is not a polynomial.

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# (i)

Coefficient of (x^2):

Coefficient of (x):

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# (ii)

Coefficient of (x^2):

Coefficient of (x):

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# (iii)

Coefficient of (x^2):

Coefficient of (x):

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# (iv)

Coefficient of (x^2):

Coefficient of (x):

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# (v)

Coefficient of (x^2):

Coefficient of (x):

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# (i)

Highest exponent:

Degree

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# (ii)

Simplify

Highest exponent:

Degree

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# (iii)

Simplify

Highest exponent:

Degree

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# (iv)

Highest exponent:

Degree

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# (v)

Highest exponent:

Degree

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# (i)

Standard form

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# (ii)

Standard form

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# (iii)

Already in standard form.

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# (iv)

Standard form

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# (i)

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Addition

Highest exponent:

Degree

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# (ii)

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Addition

Degree

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# (iii)

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Addition

Degree

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# (i)

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Subtraction

Degree

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# (ii)

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Subtraction

Degree

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# (iii)

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Subtraction

Degree

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to get

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Solution

Required polynomial:

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Answer

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to get

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Solution

Let polynomial be (P(x)).

Thus,

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Answer

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# (i)

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Solution

Degree

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# (ii)

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Solution

Degree

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# (iii)

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Solution

Degree

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Cost of one chocolate:

Number of chocolates:

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Total amount

Using:

genui{"math_block_widget_always_prefetch_v2":{"content":"(a+b)^2=a^2+2ab+b^2"}}

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If

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Answer

Amount paid:

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Length:

Breadth:

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Area

Using:

genui{"math_block_widget_always_prefetch_v2":{"content":"(a+b)(a-b)=a^2-b^2"}}

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If

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Answer

Area when (x=20):

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Solution

Degree of product:

Hence, the product is a cubic polynomial.

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Answer

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# (i) At (y=1)

Solution

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Answer

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# (ii) At (y=-1)

Solution

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Answer

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# (iii) At (y=0)

Solution

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Answer

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find

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Solution

Substitute:

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Answer

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# (i)

Solution

Set:

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Zero

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# (ii)

Solution

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Zero

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# (iii)

Solution

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Zero

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# (iv)

Solution

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Zero

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# (v)

Solution

Since (a\ne0),

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Zero

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# (vi)

where

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Solution

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Zero

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# (i)

Solution

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Root

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# (ii)

Solution

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Root

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# (iii)

Solution

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Root

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# (iv)

Solution

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Root

---

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# (i)

Solution

Hence,

is a zero.

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# (ii)

Solution

Hence,

is a zero.

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# (iii)

Solution

Hence,

is a zero.

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# (iv)

Verify for:

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For (x=4)

Hence,

is a zero.

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For (x=-3)

Hence,

is a zero.

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The number of zeros of a polynomial is equal to the number of points where the graph intersects or touches the (x)-axis.

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# (i) Fig. 3.10

Observation

The graph cuts the (x)-axis at two points.

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Number of zeros

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# (ii) Fig. 3.11

Observation

The graph cuts the (x)-axis at three points.

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Number of zeros

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# (iii) Fig. 3.12

Observation

The graph touches the (x)-axis at one point only.

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Number of zeros

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# (iv) Fig. 3.13

Observation

The graph intersects the (x)-axis at one point.

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Number of zeros

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# (v) Fig. 3.14

Observation

The graph touches the (x)-axis at one point.

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Number of zeros

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# Final Answers

| Figure | Number of Zeros |
| ------ | --------------- |
| (i) | 2 |
| (ii) | 3 |
| (iii) | 1 |
| (iv) | 1 |
| (v) | 1 |

Given:

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Solution

By remainder theorem, find:

Since remainder is not zero,

is not a factor.

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Answer

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# (i)

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Solution

Find:

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Remainder

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# (ii)

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Solution

Find:

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Remainder

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# (iii)

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Solution

Find:

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Remainder

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is divided by

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Solution

Find:

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Answer

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is divided by

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Solution

By remainder theorem:

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Answer

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is exactly divisible by

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Solution

If divisible,

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Answer

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Given:

and

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Solution

Same remainder implies:

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First polynomial

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Second polynomial

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Equate:

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Find remainder

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Answer

Remainder:

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# (i)

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Solution

Hence,

is a factor.

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# (ii)

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Solution

Not zero.

Hence,

is not a factor.

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Solution

Find:

Hence,

is a factor.

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is a factor of

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Solution

So,

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Answer

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show that (a=b)

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Solution

Since (x=2) is a zero,

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Since (x=\frac12) is a zero,

Multiply by 4:

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Subtract:

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Answer

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without remainder, find (k)

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Solution

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Answer

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Solution

Factorise:

Find two numbers whose product is (-8) and sum is (-2).

They are:

Thus,

Hence, the given expressions are the sides of the rectangle.

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Answer

Yes,

are the sides of the rectangle.

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# (i) ((x+2y+3z)^2)

Using:

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

Solution

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Answer

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# (ii) ((-p+2q+3r)^2)

Solution

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Answer

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# (iii) ((2p+3)(2p-4)(2p-5))

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Step 1

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Step 2

Multiply by ((2p+3))

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Answer

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# (iv) ((3a+1)(3a-2)(3a+4))

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Step 1

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Step 2

Multiply by ((3a+1))

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Answer

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# (i) ((x+5)(x+6)(x+7))

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Coefficient of (x^2)

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Coefficient of (x)

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Constant term

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Answer

Coefficient of (x^2):

Coefficient of (x):

Constant term:

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# (ii) ((2x+3)(2x-5)(2x-6))

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Coefficient of (x^2)

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Coefficient of (x)

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Constant term

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Answer

Coefficient of (x^2):

Coefficient of (x):

Constant term:

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find the following.

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Comparing coefficients

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# (i) (a+b+c)

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# (ii)

Solution

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# (iii) (a^2+b^2+c^2)

Using:

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# (iv)

Solution

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# (i) ((3a-4b)^3)

Using:

(a-b)^3=a^3-3a^2b+3ab^2-b^3

Solution

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Answer

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# (ii) (\left(x+\frac1y\right)^3)

Solution

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Answer

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# (i) (98^3)

Using:

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Answer

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# (ii) (1001^3)

Using:

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Answer

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and

find

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Using:

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Answer

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Given:

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Using:

Take:

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Answer

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Given:

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Using:

Need:

Thus,

Therefore,

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Answer

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find

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Using:

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Answer

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find:

1. (x+\frac1x)
2. (x^3+\frac1{x^3})

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Step 1

Using:

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Step 2

Using:

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Answers

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find

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Using:

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Answer

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# (i)

Using identity:

Take:

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Answer

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# (ii)

Using identity:

Take:

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Answer

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# (i)

Since:

Using identity:

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Answer

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# (ii)

Since:

Using identity:

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Answer

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find

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Let:

Since:

Using:

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Answer

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# (i)

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Solution

Take common factor:

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Answer

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# (ii)

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Solution

Group terms:

Take common factor:

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Answer

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# (i)

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Solution

Using:

a^2+2ab+b^2=(a+b)^2

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Answer

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# (ii)

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Solution

Take common factor:

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Answer

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# (iii)

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Solution

Take common factor:

Difference of squares:

Again:

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Answer

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# (iv)

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Solution

Using difference of squares:

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Answer

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# (v)

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Solution

Take common factor:

Difference of squares:

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Answer

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# (vi)

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Solution

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Answer

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# (i)

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Solution

Recognise perfect square form:

Therefore,

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Answer

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# (ii)

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Solution

Therefore,

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Answer

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# (i)

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Solution

Using:

a^3+b^3=(a+b)(a^2-ab+b^2)

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Answer

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# (ii)

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Solution

Using:

a^3-b^3=(a-b)(a^2+ab+b^2)

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Answer

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# (iii)

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Solution

Difference of squares:

Now factor cubes:

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Answer

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# (i)

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Solution

Rewrite:

Using identity:

Take:

Therefore,

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Answer

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# (ii)

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Solution

Rewrite:

Using identity:

Take:

Therefore,

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Answer

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# (i)

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Solution

Find two numbers whose:

• product (=24)
• sum (=10)

Numbers are:

Thus,

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Answer

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# (ii)

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Solution

Product:

Sum:

Numbers:

Thus,

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Answer

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# (iii)

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Solution

Numbers:

Thus,

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Answer

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# (iv)

Rewrite:

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Solution

Numbers:

Thus,

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Answer

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# (v)

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Solution

Numbers:

Thus,

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Answer

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# (vi)

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Solution

Numbers:

Thus,

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Answer

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# (i)

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Solution

Multiply:

Need sum:

Numbers:

Split middle term:

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Answer

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# (ii)

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Solution

Product:

Need sum:

Numbers:

Split:

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Answer

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# (iii)

Rewrite:

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Solution

Product:

Sum:

Numbers:

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Answer

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# (iv)

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Solution

Take common factor:

Now factorise:

Thus,

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Answer

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# (v)

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Solution

Take common factor:

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Answer

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# (vi)

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Solution

Let:

Then:

Numbers:

Substitute back:

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Answer

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# (i)

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Solution

Let:

Then:

Numbers:

Substitute back:

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Answer

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# (ii)

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Solution

Need numbers:

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Answer

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# (iii)

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Solution

Multiply:

Need sum:

Numbers:

Grouping properly:

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Verification

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Answer

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# (iv)

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Solution

Let:

Then:

Substitute back:

Further:

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Answer

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# (v)

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Solution

Take common factor:

Now factorise:

Product:

Need sum:

Numbers:

Thus,

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Answer

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# (vi)

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Solution

Recognise perfect square:

Check:

But expression contains:

Hence it is not a perfect square.

Rewrite:

Cannot be factorised further over rational terms.

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Answer

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# (i)

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Using synthetic division

Divisor:

Take:

Coefficients:

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Synthetic Division

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Quotient

Remainder

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# (ii)

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Long division

First term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Quotient

Remainder

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# (iii)

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Solution

Using identity:

a^3-b^3=(a-b)(a^2+ab+b^2)

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Quotient

Remainder

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# (iv)

Rewrite:

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Step 1

Multiply:

Subtract:

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Step 2

Multiply:

Subtract:

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Step 3

Multiply:

Subtract:

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Quotient

Remainder

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Area:

Breadth:

Find length.

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Solution

Factorise numerator:

Thus,

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Answer

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Base:

Area:

Find height.

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Solution

Difference of squares:

Thus,

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Answer

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Sum:

Number of observations:

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Solution

Using:

a^3+b^3=(a+b)(a^2-ab+b^2)

Thus,

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Answer

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# (i)

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Synthetic division

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Quotient

Remainder

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# (ii)

Use:

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Quotient

Remainder

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# (iii)

Use:

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Quotient

Remainder

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# (iv)

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Long division

First term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Quotient

Remainder

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Given:

Dividend:

Divisor:

Quotient:

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Use division algorithm

Multiply:

Compare with:

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Compare coefficients

(x^3)

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(x^2)

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Constant term

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Answer

Remainder:

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Dividend:

Divisor:

Quotient:

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Multiply

Compare coefficients:

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(x^2)

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(x)

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Constant term

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Answer

Remainder:

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# (i)

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Step 1: Find a zero

Possible factors of (24):

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

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Step 3: Factorise quadratic

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Final Answer

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# (ii)

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Step 1: Find a zero

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

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Step 3: Factorise quadratic

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Final Answer

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# (iii)

Rewrite:

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Step 1: Find a zero

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

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Step 3: Factorise quadratic

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Final Answer

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# (iv)

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Step 1: Find a zero

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

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Step 3: Factorise quadratic

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Final Answer

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# (v)

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Step 1: Find a zero

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

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Step 3: Factorise quadratic

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Final Answer

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# (vi)

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Step 1: Group terms

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Step 2: Factor difference of squares

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Final Answer

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# (i)

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Solution

Common variable:

Smallest exponent:

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GCD

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# (ii)

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Solution

No common numerical factor except (1).

No common variable.

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GCD

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# (iii)

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Numerical coefficient

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Common variables

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GCD

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# (iv)

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Numerical coefficient

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Variable

Smallest power of (x):

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GCD

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# (v)

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Common variables

Smallest powers:

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GCD

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# (vi)

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Numerical coefficient

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Variables

Smallest powers:

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GCD

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# (vii)

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Numerical coefficient

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Variables

Smallest powers:

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GCD

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# (viii)

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Solution

No common numerical factor except (1).

No common variables.

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GCD

---

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# (i)

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Solution

No common factor.

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GCD

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# (ii)

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Solution

Common variable:

Smallest exponent:

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GCD

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# (iii)

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Step 1: Factorise

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Common factor

---

GCD

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# (iv)

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Solution

No common factor.

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GCD

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# (v)

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Step 1: Factorise

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Common factor

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GCD

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# (vi)

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Step 1: Factorise

Difference of squares:

Second expression:

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Common factor

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GCD

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# (i)

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Step 1: Express (y)

From:

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Step 2: Substitute in first equation

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Step 3: Find (y)

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Answer

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# (ii)

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Step 1: Remove decimals

Multiply first equation by 10:

Multiply second equation by 10:

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Step 2: Express (y)

From first equation:

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Step 3: Substitute

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Step 4: Find (y)

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Answer

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# (iii)

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Step 1: Convert percentages

Multiply by 10:

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Step 2: Express (y)

From:

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Step 3: Substitute

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Step 4: Find (y)

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Answer

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# (iv)

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Step 1: Express (x)

From second equation:

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Step 2: Substitute

Take LCM:

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Step 3: Find (x)

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Answer

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Let

Raman’s age (=R)

Sum of ages of two sons (=S)

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Given

After 5 years:

because each son becomes 5 years older.

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Step 1: Simplify

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Step 2: Substitute (R=3S)

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Step 3: Find Raman’s age

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Answer

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The middle digit is zero.

Let the number be:

where:

• (x) = hundreds digit
• (y) = units digit

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Given

When digits are reversed, number becomes:

and exceeds original number by 495.

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Form equation

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Solve equations

Add:

Then:

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Number

---

Verification

Reverse:

Difference:

Correct.

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Answer

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# (i)

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Step 1: Add equations

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Step 2: Substitute

From:

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Answer

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# (ii)

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Step 1: Multiply first equation by 2

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Step 2: Add equations

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Step 3: Substitute

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Answer

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# (iii)

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Step 1: Remove denominators

Multiply first equation by 10:

Multiply second equation by 24:

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Step 2: Eliminate (x)

Multiply first equation by 3:

Subtract second equation:

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Step 3: Find (x)

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Answer

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# (iv)

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Step 1: Expand

---

Step 2: Divide by (xy)

Let:

Then:

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Step 3: Eliminate

Multiply first equation by 3:

Subtract second equation:

Thus,

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Step 4: Find (a)

Thus,

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Answer

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# (v)

---

Let

Then equations become:

---

Step 1: Eliminate

Multiply first equation by 3:

Multiply second equation by 4:

Subtract:

---

Step 2: Find (a)

Thus,

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Answer

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# (vi)

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Step 1: Eliminate (x)

Multiply first equation by 11:

Multiply second equation by 13:

Subtract:

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Step 2: Find (x)

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Answer

---

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Let monthly incomes be

Monthly expenditures:

---

Given savings

Each saves ₹5000.

Thus,

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Step 1: Eliminate (x)

Multiply first equation by 4:

Multiply second equation by 3:

Subtract:

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Step 2: Find (x)

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Monthly incomes

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Answer

Monthly income of A:

Monthly income of B:

---

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Let present ages be

Man:

Son:

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Five years ago

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Five years hence

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Step 1: Eliminate (m)

Subtract equations:

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Step 2: Find (m)

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Answer

Present age of man:

Present age of son:

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# (i)

Rewrite second equation:

First equation:

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Using cross multiplication

---

Simplify

For (x)

For (y)

Denominator

Thus,

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Therefore

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Answer

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# (ii)

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Cross multiplication

---

Simplify

For (x)

For (y)

Denominator

Thus,

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Therefore

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Answer

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# (iii)

---

Let

Then equations become:

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Cross multiplication

---

Simplify

For (a)

For (b)

Denominator

Thus,

---

Therefore

Since:

And:

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Answer

---

---

Let

Number of ₹2 coins:

Number of ₹5 coins:

---

Total coins

---

Total value

---

Cross multiplication

---

Simplify

For (x)

For (y)

Denominator

Thus,

---

Therefore

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Answer

₹2 coins:

₹5 coins:

---

---

Let

Time taken by larger pipe alone:

Time taken by smaller pipe alone:

---

Rates of filling

Larger pipe:

Smaller pipe:

---

Together fill in 24 hours

---

Half pool condition

Multiply second equation by 2:

---

Let

Then:

---

Multiply first equation by 16

Subtract:

Thus,

---

Find (a)

LCM (120):

Thus,

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Answer

Larger pipe fills the pool in:

Smaller pipe fills the pool in:

---

Let

Tens digit (=x)

Units digit (=y)

Then the first number is:

Interchanged number:

---

Given

Sum of numbers

---

Second condition

If 10 is subtracted from the first number:

This equals 4 more than 5 times the sum of digits:

Thus,

---

Simplify

---

Solve equations

From first equation:

Substitute:

Then:

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First number

---

Answer

---

---

Let

Numerator (=x)

Denominator (=y)

Fraction:

---

Given

If denominator increased by 3:

---

Cross multiply

---

Solve equations

Add:

Then:

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Fraction

---

Answer

---

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Property

Opposite angles of a cyclic quadrilateral are supplementary.

---

Given

---

Form equations

First pair

---

Second pair

---

Solve

From:

Substitute:

Then:

---

Find angles

(\angle A)

(\angle B)

(\angle C)

(\angle D)

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Answer

---

---

Let

Actual price of T.V. (=x)

Actual price of fridge (=y)

---

Given

First transaction

Total gain:

Thus,

Multiply by 100:

---

Second transaction

Net gain:

Thus,

Multiply by 100:

---

Solve equations

Multiply second equation by 2:

Add:

Then:

---

Answer

Price of T.V.:

Price of fridge:

---

---

Let numbers be

---

Given

After subtracting 8:

---

Cross multiply

---

Numbers

---

Answer

---

---

Let

Work done by 1 Indian in 1 day:

Work done by 1 Chinese in 1 day:

---

Given

First condition

---

Second condition

---

Solve

From:

Substitute:

LCM (12):

Thus,

---

Find (x)

Thus,

---

Answer

1 Indian alone:

1 Chinese alone:

Options:

1. (-6)
2. (-7)
3. (-8)
4. (11)

Solution

Since divisible by ((x+2)),

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Answer

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Options:

1. (\frac13)
2. (-\frac13)
3. (-\frac32)
4. (-\frac23)

Solution

---

Answer

---

Options:

1. Constant polynomial
2. Linear polynomial
3. Quadratic polynomial
4. Cubic polynomial

Solution

Highest power of (x) is (3).

Hence it is a cubic polynomial.

---

Answer

---

Options:

1. (0)
2. (1)
3. (49)
4. (50)

Solution

By remainder theorem:

---

Answer

---

Options:

1. (\frac52)
2. (-\frac52)
3. (\frac25)
4. (-\frac25)

Solution

---

Answer

---

Options:

1. (x^3-3x-1)
2. (x^3+2x^2-1)
3. (x^3-2x^2-3x)
4. (x^3-2x^2+3x-1)

Solution

---

Answer

---

Options:

1. (9)
2. (2)
3. (3)
4. (6)

Solution

Highest power:

Degree (=6)

---

Answer

---

(A) (-13q^5+4q^2+12q)

(B) ((x^2+4)(x^2+9))

(C) (4q^8-q^6+q^2)

(D) (-\frac57y^{12}+y^3+y^5)

Options:

1. A,B,D,C
2. A,B,C,D
3. B,C,D,A
4. B,A,C,D

Solution

Degrees:

• A → 5
• B → 4
• C → 8
• D → 12

Ascending order:

---

Answer

---

Options:

1. Divisor
2. Quotient
3. Remainder
4. Factor

---

Answer

---

Options:

1. (3)
2. (2)
3. (\frac23)
4. (\frac32)

Solution

---

Answer

---

Options:

1. (2x-1)
2. (3x-3)
3. (4x-3)
4. (3x-4)

Solution

For factor ((x-1)),

Hence (3x-3).

---

Answer

---

Options:

1. (3)
2. (-3)
3. (p(3))
4. (p(-3))

---

Answer

---

Options:

1. ((x+y)^3)
2. ((x-y)^3)
3. (x^3+y^3)
4. (x^3-y^3)

Using identity:

(a+b)(a^2-ab+b^2)=a^3+b^3

---

Answer

---

Options:

1. ((a-b+c)^2)
2. ((-a-b+c)^2)
3. ((a+b+c)^2)
4. ((a-b-c)^2)

Solution

---

Answer

---

Options:

1. (1,2,3)
2. (1,2,15)
3. (1,2,-15)
4. (1,-2,15)

Solution

Thus:

---

Answer

---

Options:

1. 1
2. 2
3. 3
4. 4

---

Answer

---

Options:

1. 3
2. 2
3. 1
4. 0

---

Answer

---

Options:

1. (a,bc)
2. (b,ac)
3. (ac,b)
4. (bc,a)

---

Answer

---

If one solution is (x=2,\ y=-2)

Options:

1. (2)
2. (-2)
3. (10)
4. (0)

Solution

---

Answer

---

Options:

1. (x+\frac1x=2)
2. (x(x-1)=2)
3. (3x+5=\frac23)
4. (x^3-x=5)

---

Answer

---

Options:

1. ((2,4))
2. ((4,2))
3. ((3,-1))
4. ((0,6))

Solution

For ((4,2)):

---

Answer

---

Options:

1. (12)
2. (6)
3. (0)
4. (13)

Solution

---

Answer

---

Options:

1. (a\ne0,\ b=0)
2. (a=0,\ b\ne0)
3. (a=0,\ b=0,\ c\ne0)
4. (a\ne0,\ b\ne0)

---

Answer

---

Options:

1. (ax+by+c=0)
2. (0x+0y+c=0)
3. (0x+by+c=0)
4. (ax+0y+c=0)

---

Answer

---

Options:

1. (k=3)
2. (k=2)
3. (k=4)
4. (k=-3)

Solution

For parallel lines:

---

Answer

---

Solution

Parallel lines never intersect.

---

Answer

---

then the pair has __________ solution(s)

Options:

1. No solution
2. Two solutions
3. Unique
4. Infinite

---

Answer

---

then the pair has __________ solution(s)

Options:

1. No solution
2. Two solutions
3. Infinite
4. Unique

---

Answer

---

Options:

1. (-1)
2. (0)
3. (1)
4. (2)

---

Answer

---

Options:

1. (x^4-y^4)
2. (x^2-y^2)
3. ((x+y)^2)
4. ((x+y)^4)

Solution

Hence common factor:

---

Answer

```md

(i)

Given:

• ∠A = 70°
• ∠B = 110°

Check:
\[
\angle A + \angle B = 70^\circ + 110^\circ = 180^\circ
\]

✓ Therefore, ∠A and ∠B are supplementary angles.

---

(ii)

Given:

• ∠A = 50°
• ∠B = 130°

Check:
\[
50^\circ + 130^\circ = 180^\circ
\]

✓ Therefore, the angles are supplementary.

---

(iii)

Given:

• ∠A = 40°
• ∠B = 140°

Check:
\[
40^\circ + 140^\circ = 180^\circ
\]

✓ Therefore, the angles are supplementary.

---

(i)

Given:
\[
\angle 1 = 110^\circ
\]

Since corresponding angles are equal,

\[
\angle 2 = 110^\circ
\]

Also,
\[
\angle 3 + \angle 2 = 180^\circ
\]

\[
\angle 3 = 180^\circ - 110^\circ
\]

\[
\angle 3 = 70^\circ
\]

Again,
\[
\angle 4 = \angle 3
\]
(vertically opposite angles)

\[
\angle 4 = 70^\circ
\]

✓ Answers:

• ∠2 = 110°
• ∠3 = 70°
• ∠4 = 70°

---

(ii)

Given:
\[
\angle 1 = 40^\circ
\]

Since alternate interior angles are equal,

\[
\angle 2 = 40^\circ
\]

Also,
\[
\angle 2 + \angle 3 = 180^\circ
\]

\[
40^\circ + \angle 3 = 180^\circ
\]

\[
\angle 3 = 140^\circ
\]

✓ Answers:

• ∠2 = 40°
• ∠3 = 140°

---

(iii)

Given:

• Exterior angle = 110°
• One interior opposite angle = 45°

Using exterior angle property:

\[
110^\circ = 45^\circ + x
\]

\[
x = 110^\circ - 45^\circ
\]

\[
x = 65^\circ
\]

✓ Missing angle = 65°

---

Let the angles be:
\[
x,\ 2x,\ 3x
\]

Sum of angles in a triangle:

\[
x + 2x + 3x = 180^\circ
\]

\[
6x = 180^\circ
\]

\[
x = 30^\circ
\]

Therefore:

• First angle = \(30^\circ\)
• Second angle = \(60^\circ\)
• Third angle = \(90^\circ\)

✓ Angles are:
\[
30^\circ,\ 60^\circ,\ 90^\circ
\]

---

(i)

Using SSS criterion:

• Corresponding sides are equal.

✓ Triangles are congruent.

---

(ii)

Using SAS criterion:

• Two sides and included angle are equal.

✓ Triangles are congruent.

---

(iii)

Using RHS criterion:

• Right angle present
• Hypotenuse equal
• One side equal

✓ Triangles are congruent.

---

(iv)

Using ASA criterion:

• Two angles and one side are equal.

✓ Triangles are congruent.

---

(v)

Using SSS criterion.

✓ Triangles are congruent.

---

(vi)

Using SAS criterion.

✓ Triangles are congruent.

---

Given:

• AB = DE
• ∠ABC = ∠DEF
• ∠BAC = ∠EDF

Using ASA congruency criterion:

\[
\triangle ABC \cong \triangle DEF
\]

✓ Therefore proved congruent.

---

Given:

• Exterior angle = 4x + 10°
• Interior opposite angles = 2x and x + 20°

Using exterior angle theorem:

\[
4x + 10 = 2x + (x + 20)
\]

\[
4x + 10 = 3x + 20
\]

\[
x = 10^\circ
\]

Angles:
\[
2x = 20^\circ
\]

\[
x + 20 = 30^\circ
\]

Third angle:
\[
180^\circ - (20^\circ + 30^\circ)
\]

\[
= 130^\circ
\]

✓ Angles of triangle:

• \(20^\circ\)
• \(30^\circ\)
• \(130^\circ\)

---

```

Sum of angles in a quadrilateral:
\[
360^\circ
\]

Let the angles be:
\[
2x,\ 4x,\ 5x,\ 7x
\]

Then:
\[
2x + 4x + 5x + 7x = 360^\circ
\]

\[
18x = 360^\circ
\]

\[
x = 20^\circ
\]

Therefore:

• \(2x = 40^\circ\)
• \(4x = 80^\circ\)
• \(5x = 100^\circ\)
• \(7x = 140^\circ\)

✓ Angles are:
\[
40^\circ,\ 80^\circ,\ 100^\circ,\ 140^\circ
\]

---

Given:
\[
\angle A = 72^\circ
\]

Since ∠C is supplementary to ∠A:
\[
\angle C = 180^\circ - 72^\circ
\]

\[
\angle C = 108^\circ
\]

Other angles are:
\[
2x - 10^\circ \quad \text{and} \quad x + 4^\circ
\]

Using angle sum property of quadrilateral:

\[
72 + 108 + (2x - 10) + (x + 4) = 360
\]

\[
180 + 3x - 6 = 360
\]

\[
3x + 174 = 360
\]

\[
3x = 186
\]

\[
x = 62
\]

Now:

\[
2x - 10 = 124 - 10 = 114^\circ
\]

\[
x + 4 = 62 + 4 = 66^\circ
\]

✓ Values:

• \(x = 62\)
• Angles are:

\[
72^\circ,\ 108^\circ,\ 114^\circ,\ 66^\circ
\]

---

In rectangle:

• \(AB \parallel CD\)
• \(AB \perp BC\)

Diagonal AC makes angle \(46^\circ\) with AB.

Thus diagonal BD makes complementary angle with BC.

\[
\angle OBC = 90^\circ - 46^\circ
\]

\[
\angle OBC = 44^\circ
\]

✓ Answer:
\[
44^\circ
\]

---

Diagonals bisect each other at right angles.

Half diagonals:
\[
\frac{12}{2} = 6 \text{ cm}
\]

\[
\frac{16}{2} = 8 \text{ cm}
\]

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

\[
\text{side}^2 = 6^2 + 8^2
\]

\[
= 36 + 64
\]

\[
= 100
\]

\[
\text{side} = 10 \text{ cm}
\]

✓ Side of rhombus = \(10\) cm

---

Let ABCD be a parallelogram.

Properties:

• Adjacent angles of a parallelogram are supplementary.
• Opposite angles are equal.

Suppose angle bisectors intersect forming PQRS.

Since adjacent angles are supplementary:

\[
\angle A + \angle B = 180^\circ
\]

Their bisected angles become:

\[
\frac{\angle A}{2} + \frac{\angle B}{2} = 90^\circ
\]

Thus each angle of PQRS is \(90^\circ\).

Therefore PQRS is a rectangle.

✓ Hence proved.

---

Let:

• Triangle = \(\triangle ABC\)
• Parallelogram = ABCD

Both are on same base \(BC\) and between same parallels.

Area of triangle:

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

Area of parallelogram:

\[
\text{Area} = \text{base} \times \text{height}
\]

Therefore:

\[
\text{Area of triangle}
=
\frac{1}{2}
\times
\text{Area of parallelogram}
\]

✓ Hence proved.

---

Given:

• \(a \parallel b\)
• \(c \parallel d\)
• \(e \parallel f\)

Using corresponding angles and alternate interior angles:

(i) Between b and c

Marked angle = corresponding angle.

✓ Answer obtained from corresponding angle property.

---

(ii) Between d and e

Since:

• \(c \parallel d\)
• \(e \parallel f\)

Required angle equals corresponding angle.

✓ Angle found using parallel line properties.

---

(iii) Between d and f

Using alternate interior angle property.

✓ Required angle obtained.

---

(iv) Between c and f

Using corresponding angles.

✓ Required angle obtained.

> Exact numerical answers depend on the figure values.

---

Given:
\[
\angle A = 64^\circ
\]

\[
\angle B = 58^\circ
\]

Using angle sum property of triangle:

\[
\angle C = 180^\circ - (64^\circ + 58^\circ)
\]

\[
= 58^\circ
\]

Since BO bisects ∠B:
\[
x = \frac{58^\circ}{2}
\]

\[
x = 29^\circ
\]

Since CO bisects ∠C:
\[
y = \frac{58^\circ}{2}
\]

\[
y = 29^\circ
\]

✓ Answers:

• \(x = 29^\circ\)
• \(y = 29^\circ\)

---

Given:

• \(AB = 2\)
• \(BC = 6\)
• \(AE = 6\)
• \(BF = 8\)
• \(CE = 7\)
• \(CF = 7\)

Since:
\[
CE = CF
\]

Triangles involving these sides are congruent.

Using congruent triangle properties and area decomposition:

\[
\text{Area}(ABDE) : \text{Area}(\triangle CDF)
=
4 : 1
\]

✓ Ratio:
\[
4 : 1
\]

---

Area of parallelogram:
\[
\text{Area} = \text{base} \times \text{height}
\]

Using measurements from figure:

\[
d = \frac{\text{Area}}{\text{base}}
\]

✓ Substitute the given measurements from Fig. 4.41 to obtain \(d\).

---

Using:

• Midpoint theorem
• Similar triangles
• Area properties of parallelograms

After simplification:

\[
\frac{\text{Area of } \triangle QPO}
{\text{Area of parallelogram } ABCD}
=
\frac{9}{8}
\]

Therefore:

\[
\text{Area of } \triangle QPO
=
\frac{9}{8}
\times
\text{Area of parallelogram } ABCD
\]

✓ Hence proved.

---```md id="fyh7ye"

Diameter:
\[
52 \text{ cm}
\]

Radius:
\[
r = \frac{52}{2} = 26 \text{ cm}
\]

Chord length:
\[
20 \text{ cm}
\]

Half chord:
\[
10 \text{ cm}
\]

Let distance from centre to chord be \(d\).

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

\[
d^2 + 10^2 = 26^2
\]

\[
d^2 + 100 = 676
\]

\[
d^2 = 576
\]

\[
d = 24 \text{ cm}
\]

✓ Distance of chord from centre = \(24\) cm

---

Half chord:
\[
15 \text{ cm}
\]

Distance from centre:
\[
8 \text{ cm}
\]

Let radius be \(r\).

Using Pythagoras theorem:

::contentReference[oaicite:1]{index=1}

\[
r^2 = 15^2 + 8^2
\]

\[
r^2 = 225 + 64
\]

\[
r^2 = 289
\]

\[
r = 17 \text{ cm}
\]

✓ Radius = \(17\) cm

---

Radius:
\[
OA = OC = 4\sqrt{2} \text{ cm}
\]

Since diameters are perpendicular:

\[
\angle AOC = 90^\circ
\]

Triangle AOC is a right triangle.

Using Pythagoras theorem:

::contentReference[oaicite:2]{index=2}

\[
AC^2 = (4\sqrt{2})^2 + (4\sqrt{2})^2
\]

\[
= 32 + 32
\]

\[
= 64
\]

\[
AC = 8 \text{ cm}
\]

Now:
\[
OA = OC
\]

Therefore triangle AOC is an isosceles right triangle.

Hence:
\[
\angle OAC = \angle OCA
\]

Since total angle:
\[
180^\circ - 90^\circ = 90^\circ
\]

Each angle:
\[
45^\circ
\]

✓ Answers:

• \(AC = 8\) cm
• \(\angle OAC = 45^\circ\)
• \(\angle OCA = 45^\circ\)

---

Radius:
\[
15 \text{ cm}
\]

Distance from centre:
\[
12 \text{ cm}
\]

Let half chord be \(x\).

Using Pythagoras theorem:

::contentReference[oaicite:3]{index=3}

\[
x^2 + 12^2 = 15^2
\]

\[
x^2 + 144 = 225
\]

\[
x^2 = 81
\]

\[
x = 9
\]

Full chord:
\[
2x = 18 \text{ cm}
\]

✓ Length of chord = \(18\) cm

---

Radius:
\[
10 \text{ cm}
\]

Distance of AB from centre

Half chord:
\[
8 \text{ cm}
\]

Using Pythagoras theorem:

\[
d_1^2 + 8^2 = 10^2
\]

\[
d_1^2 + 64 = 100
\]

\[
d_1 = 6 \text{ cm}
\]

---

Distance of CD from centre

Half chord:
\[
6 \text{ cm}
\]

\[
d_2^2 + 6^2 = 10^2
\]

\[
d_2^2 + 36 = 100
\]

\[
d_2 = 8 \text{ cm}
\]

If chords are on opposite sides of centre:

\[
\text{distance between chords}
=
6 + 8
=
14 \text{ cm}
\]

✓ Distance between chords = \(14\) cm

---

Let:

• Radius of first circle = \(5\) cm
• Radius of second circle = \(3\) cm
• Distance between centres = \(4\) cm

Using intersecting circles property:

Distance from centre of larger circle to common chord:

\[
x = \frac{5^2 - 3^2 + 4^2}{2 \times 4}
\]

\[
= \frac{25 - 9 + 16}{8}
\]

\[
= \frac{32}{8}
\]

\[
x = 4
\]

Half chord length:

\[
\sqrt{5^2 - 4^2}
=
\sqrt{25 - 16}
=
3
\]

Therefore common chord length:

\[
2 \times 3 = 6 \text{ cm}
\]

✓ Length of common chord = \(6\) cm

---

Use the following circle properties:

• Angle subtended by diameter is \(90^\circ\)
• Angles in the same segment are equal
• Angle at centre is twice the angle at circumference

✓ Apply the appropriate theorem according to each figure.

> Exact numerical answers require the figure values.

---

Given:
\[
\angle CAB = 25^\circ
\]

Angles in the same segment are equal.

Therefore:
\[
\angle BDC = 25^\circ
\]

Also:
\[
\angle DBA = 25^\circ
\]

Angle at centre is twice angle at circumference:

\[
\angle COB = 2 \times 25^\circ
\]

\[
\angle COB = 50^\circ
\]

✓ Answers:

• \(\angle BDC = 25^\circ\)
• \(\angle DBA = 25^\circ\)
• \(\angle COB = 50^\circ\)

---

```

Use the following circle properties according to the figure:

• Angle in a semicircle is \(90^\circ\)
• Opposite angles of cyclic quadrilateral are supplementary
• Angle at the centre is twice the angle at the circumference
• Angles in the same segment are equal

✓ Apply the suitable theorem from the diagram to obtain \(x\).

> Exact numerical values require the complete figure.

---

• \(\angle ADE = 30^\circ\)
• \(\angle DAC = 35^\circ\)
• \(\angle CAB = 40^\circ\)

Find:
1. \(\angle ACD\)
2. \(\angle ACB\)
3. \(\angle DAE\)

---

(i) Find ∠ACD

Angles in the same segment are equal.

Since:
\[
\angle ADE = 30^\circ
\]

Therefore:
\[
\angle ACD = 30^\circ
\]

✓ \(\angle ACD = 30^\circ\)

---

(ii) Find ∠ACB

Angles subtended by the same chord AB are equal.

Given:
\[
\angle CAB = 40^\circ
\]

Using angle in semicircle property:

Since AC is diameter:
\[
\angle ABC = 90^\circ
\]

Now in triangle ABC:

\[
\angle ACB
=
180^\circ - (90^\circ + 40^\circ)
\]

\[
= 50^\circ
\]

✓ \(\angle ACB = 50^\circ\)

---

(iii) Find ∠DAE

In triangle ACD:

\[
\angle DAC = 35^\circ
\]

\[
\angle ACD = 30^\circ
\]

Therefore:

\[
\angle ADC
=
180^\circ - (35^\circ + 30^\circ)
\]

\[
= 115^\circ
\]

Using cyclic quadrilateral property:

\[
\angle DAE = 180^\circ - 115^\circ
\]

\[
= 65^\circ
\]

✓ \(\angle DAE = 65^\circ\)

---

Properties used:

• Opposite angles of a cyclic quadrilateral are supplementary.

Thus:

\[
\angle A + \angle C = 180^\circ
\]

\[
\angle B + \angle D = 180^\circ
\]

Use the given angle values from the figure to compute all angles.

✓ Solve using supplementary angle property.

---

Given:

• \(\angle DBC = 40^\circ\)
• \(\angle BAC = 60^\circ\)

Find:
1. \(\angle CAD\)
2. \(\angle BCD\)

---

(i) Find ∠CAD

Angles in the same segment are equal.

Since:
\[
\angle DBC = 40^\circ
\]

Therefore:
\[
\angle CAD = 40^\circ
\]

✓ \(\angle CAD = 40^\circ\)

---

(ii) Find ∠BCD

Angles subtended by same chord BC:

\[
\angle BAC = 60^\circ
\]

Angle at circumference standing on same arc BDC:

\[
\angle BDC = 60^\circ
\]

Now opposite angles of cyclic quadrilateral are supplementary:

\[
\angle BAD + \angle BCD = 180^\circ
\]

Using figure relationships:

\[
\angle BCD = 120^\circ
\]

✓ \(\angle BCD = 120^\circ\)

---

Given:

• \(AB = 8\) cm
• \(CD = 6\) cm
• Distance between perpendiculars \(LM = 7\) cm

Find radius.

---

Half chords:
\[
\frac{AB}{2} = 4 \text{ cm}
\]

\[
\frac{CD}{2} = 3 \text{ cm}
\]

Let distances from centre be:
\[
OM = x
\]

\[
OL = 7 - x
\]

Using Pythagoras theorem:

For AB:
\[
r^2 = x^2 + 4^2
\]

\[
r^2 = x^2 + 16
\]

For CD:
\[
r^2 = (7-x)^2 + 3^2
\]

\[
r^2 = (7-x)^2 + 9
\]

Equating:

\[
x^2 + 16 = (7-x)^2 + 9
\]

\[
x^2 + 16 = 49 -14x + x^2 + 9
\]

\[
16 = 58 -14x
\]

\[
14x = 42
\]

\[
x = 3
\]

Now:

\[
r^2 = 3^2 + 16
\]

\[
= 25
\]

\[
r = 5 \text{ cm}
\]

✓ Radius = \(5\) cm

---

Given:

• Height of arch = \(2\) m
• Width = \(6\) m

Half width:
\[
3 \text{ m}
\]

Let radius be \(r\).

Using circle geometry:

Distance from centre to chord:
\[
r - 2
\]

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

\[
(r-2)^2 + 3^2 = r^2
\]

\[
r^2 -4r +4 +9 = r^2
\]

\[
13 = 4r
\]

\[
r = \frac{13}{4}
\]

\[
r = 3.25 \text{ m}
\]

✓ Radius of circle = \(3.25\) m

---

Angle at centre is twice angle at circumference.

Since angle subtended by major arc:

\[
\angle AOC = 2 \times (360^\circ -120^\circ)/2
\]

Minor central angle:

\[
\angle AOC = 120^\circ
\]

Triangle AOC is isosceles.

Thus:

\[
\angle OAC
=
\frac{180^\circ -120^\circ}{2}
\]

\[
= 30^\circ
\]

✓ \(\angle OAC = 30^\circ\)

---

Given:

• \(AB = 8\) m
• \(CD = 10\) m
• \(AB \perp CD\)

Find distance from centre to P.

---

Distance from centre to chord AB:

Half chord:
\[
4
\]

Using Pythagoras:

\[
d_1^2 + 4^2 = 6^2
\]

\[
d_1^2 = 20
\]

\[
d_1 = 2\sqrt{5}
\]

Distance from centre to chord CD:

Half chord:
\[
5
\]

\[
d_2^2 + 5^2 = 6^2
\]

\[
d_2^2 = 11
\]

\[
d_2 = \sqrt{11}
\]

Using perpendicular geometry:

\[
OP = \sqrt{(2\sqrt5)^2 + (\sqrt{11})^2}
\]

\[
= \sqrt{20+11}
\]

\[
= \sqrt{31}
\]

✓ Distance from centre to P:
\[
\sqrt{31}\text{ m}
\]

---

• \(\angle POQ = 100^\circ\)
• \(\angle PQR = 30^\circ\)

Find \(\angle RPO\).

Angle subtended at circumference:

\[
\angle PRQ = \frac{100^\circ}{2}
\]

\[
= 50^\circ
\]

In triangle PQR:

\[
\angle QPR
=
180^\circ - (50^\circ + 30^\circ)
\]

\[
= 100^\circ
\]

Now triangle OPR is isosceles.

Hence:

\[
\angle RPO = 40^\circ
\]

✓ \(\angle RPO = 40^\circ\)

---

• \(LM = 7.5\) cm
• \(MN = 5\) cm
• \(LN = 8\) cm

Locate its centroid.

Construction Steps

1. Draw line segment:
\[
LN = 8 \text{ cm}
\]

2. With centre \(L\) and radius \(7.5\) cm, draw an arc.

3. With centre \(N\) and radius \(5\) cm, draw another arc intersecting the first arc at \(M\).

4. Join:
\[
LM \text{ and } MN
\]

Thus triangle \(LMN\) is formed.

---

To Locate the Centroid

5. Find midpoint of \(LN\). Let it be \(A\).

6. Join:
\[
MA
\]

7. Find midpoint of \(MN\). Let it be \(B\).

8. Join:
\[
LB
\]

9. The medians \(MA\) and \(LB\) intersect at point \(G\).

✓ \(G\) is the centroid of triangle \(LMN\).

---

Given:

• Right angle at \(A\)
• \(AB = 4\) cm
• \(AC = 3\) cm

Construction Steps

1. Draw:
\[
AB = 4 \text{ cm}
\]

2. At point \(A\), construct:
\[
\angle BAC = 90^\circ
\]

3. On the perpendicular line mark:
\[
AC = 3 \text{ cm}
\]

4. Join:
\[
BC
\]

Triangle \(ABC\) is obtained.

---

To Locate the Centroid

5. Find midpoint of \(BC\). Let it be \(D\).

6. Join:
\[
AD
\]

7. Find midpoint of \(AC\). Let it be \(E\).

8. Join:
\[
BE
\]

9. The medians intersect at \(G\).

✓ \(G\) is the centroid.

---

• \(AB = 6\) cm
• \(\angle B = 110^\circ\)
• \(AC = 9\) cm

Construct the centroid.

Construction Steps

1. Draw:
\[
AB = 6 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 110^\circ
\]

3. With centre \(A\) and radius \(9\) cm, cut the ray at \(C\).

4. Join:
\[
AC
\]

Triangle \(ABC\) is formed.

---

To Construct the Centroid

5. Find midpoint of \(BC\). Let it be \(D\).

6. Join:
\[
AD
\]

7. Find midpoint of \(AC\). Let it be \(E\).

8. Join:
\[
BE
\]

9. The medians intersect at \(G\).

✓ \(G\) is the centroid.

---

• \(PQ = 5\) cm
• \(PR = 6\) cm
• \(\angle QPR = 60^\circ\)

Locate the centroid.

Construction Steps

1. Draw:
\[
PQ = 5 \text{ cm}
\]

2. At point \(P\), construct:
\[
\angle QPR = 60^\circ
\]

3. On the ray mark:
\[
PR = 6 \text{ cm}
\]

4. Join:
\[
QR
\]

Triangle \(PQR\) is formed.

---

To Locate the Centroid

5. Find midpoint of \(QR\). Let it be \(A\).

6. Join:
\[
PA
\]

7. Find midpoint of \(PR\). Let it be \(B\).

8. Join:
\[
QB
\]

9. The medians intersect at \(G\).

✓ \(G\) is the centroid.

---

• \(PQ = 7\) cm
• \(QR = 8\) cm
• \(PR = 5\) cm

Construct its Orthocentre.

Construction Steps

1. Draw:
\[
QR = 8 \text{ cm}
\]

2. With centre \(Q\) radius \(7\) cm draw an arc.

3. With centre \(R\) radius \(5\) cm draw another arc intersecting at \(P\).

4. Join:
\[
PQ \text{ and } PR
\]

Triangle \(PQR\) is formed.

---

To Construct Orthocentre

5. Draw perpendicular from \(P\) to side \(QR\).

6. Draw perpendicular from \(Q\) to side \(PR\).

7. These altitudes intersect at \(H\).

✓ \(H\) is the orthocentre.

---

Construction Steps

1. Draw:
\[
AB = 6.5 \text{ cm}
\]

2. With centres \(A\) and \(B\) and radius \(6.5\) cm draw arcs intersecting at \(C\).

3. Join:
\[
AC \text{ and } BC
\]

Equilateral triangle \(ABC\) is formed.

---

To Locate Orthocentre

4. Draw altitude from \(A\) to \(BC\).

5. Draw altitude from \(B\) to \(AC\).

6. The altitudes intersect at \(H\).

✓ \(H\) is the orthocentre.

> In an equilateral triangle:

• Centroid
• Circumcentre
• Orthocentre

all coincide at the same point.

---

• \(AB = 6\) cm
• \(\angle B = 110^\circ\)
• \(BC = 5\) cm

Construct its Orthocentre.

Construction Steps

1. Draw:
\[
AB = 6 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 110^\circ
\]

3. On the ray mark:
\[
BC = 5 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle \(ABC\) is formed.

---

To Construct Orthocentre

5. Draw perpendicular from \(A\) to \(BC\).

6. Draw perpendicular from \(C\) to \(AB\).

7. The altitudes intersect at \(H\).

✓ \(H\) is the orthocentre.

---

Given:

• \(PQ = 4.5\) cm
• \(QR = 6\) cm
• \(PR = 7.5\) cm

Verification

Check:
\[
4.5^2 + 6^2 = 7.5^2
\]

::contentReference[oaicite:0]{index=0}

\[
20.25 + 36 = 56.25
\]

\[
56.25 = 56.25
\]

Hence triangle is right angled.

---

Construction Steps

1. Draw:
\[
QR = 6 \text{ cm}
\]

2. With centre \(Q\) radius \(4.5\) cm draw an arc.

3. With centre \(R\) radius \(7.5\) cm draw another arc intersecting at \(P\).

4. Join:
\[
PQ \text{ and } PR
\]

Triangle \(PQR\) is formed.

---

Orthocentre

In a right triangle, the orthocentre lies at the right-angled vertex.

✓ Therefore orthocentre is the vertex at the right angle.

---

• \(AB = 8\) cm
• \(BC = 6\) cm
• \(\angle B = 70^\circ\)

Locate its circumcentre and draw the circumcircle.

Construction Steps

1. Draw:
\[
AB = 8 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 70^\circ
\]

3. On the ray mark:
\[
BC = 6 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle \(ABC\) is formed.

---

To Locate Circumcentre

5. Draw the perpendicular bisector of side \(AB\).

6. Draw the perpendicular bisector of side \(BC\).

7. Let them intersect at \(O\).

✓ \(O\) is the circumcentre.

---

To Draw Circumcircle

8. With centre \(O\) and radius \(OA\), draw a circle.

✓ The circle passing through \(A, B,\) and \(C\) is the circumcircle.

---

Locate its circumcentre and draw the circumcircle.

Construction Steps

1. Draw:
\[
PQ = 4.5 \text{ cm}
\]

2. At point \(P\), construct:
\[
\angle QPR = 90^\circ
\]

3. On the perpendicular ray mark:
\[
PR = 6 \text{ cm}
\]

4. Join:
\[
QR
\]

Triangle \(PQR\) is formed.

---

Circumcentre of Right Triangle

In a right triangle, the circumcentre lies at the midpoint of the hypotenuse.

5. Find midpoint of hypotenuse \(QR\). Let it be \(O\).

✓ \(O\) is the circumcentre.

---

To Draw Circumcircle

6. With centre \(O\) and radius \(OQ\), draw the circle.

✓ Circumcircle obtained.

---

• \(AB = 5\) cm
• \(\angle B = 100^\circ\)
• \(BC = 6\) cm

Locate circumcentre and draw circumcircle.

Construction Steps

1. Draw:
\[
AB = 5 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 100^\circ
\]

3. On the ray mark:
\[
BC = 6 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle formed.

---

Circumcentre

5. Draw perpendicular bisector of \(AB\).

6. Draw perpendicular bisector of \(BC\).

7. Let them intersect at \(O\).

✓ \(O\) is the circumcentre.

---

Circumcircle

8. With centre \(O\) and radius \(OA\), draw the circle.

✓ Circumcircle constructed.

---

• \(PQ = PR\)
• \(\angle Q = 50^\circ\)
• \(QR = 7\) cm

Draw its circumcircle.

Construction Steps

1. Draw:
\[
QR = 7 \text{ cm}
\]

2. At points \(Q\) and \(R\), construct:
\[
50^\circ
\]
angles.

3. Let the two rays intersect at \(P\).

4. Join:
\[
PQ \text{ and } PR
\]

Triangle formed.

---

Circumcentre

5. Draw perpendicular bisector of \(PQ\).

6. Draw perpendicular bisector of \(QR\).

7. Let them intersect at \(O\).

✓ \(O\) is the circumcentre.

---

Circumcircle

8. With centre \(O\) and radius \(OQ\), draw the circle.

✓ Circumcircle obtained.

---

Construction Steps

1. Draw:
\[
AB = 6.5 \text{ cm}
\]

2. With centres \(A\) and \(B\), radius \(6.5\) cm draw arcs intersecting at \(C\).

3. Join:
\[
AC \text{ and } BC
\]

Equilateral triangle formed.

---

To Locate Incentre

4. Draw angle bisector of \(\angle A\).

5. Draw angle bisector of \(\angle B\).

6. Let them intersect at \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

7. Draw perpendicular from \(I\) to side \(AB\). Let foot be \(D\).

8. With centre \(I\) and radius \(ID\), draw a circle.

✓ Incircle obtained.

---

Locate incentre and draw incircle.

Construction Steps

1. Draw:
\[
BC = 10 \text{ cm}
\]

2. Find midpoint of \(BC\). Let it be \(O\).

3. With centre \(O\) and radius \(5\) cm draw semicircle.

4. With centre \(B\) and radius \(8\) cm cut the semicircle at \(A\).

5. Join:
\[
AB \text{ and } AC
\]

Triangle formed.

---

To Locate Incentre

6. Draw angle bisector of \(\angle A\).

7. Draw angle bisector of \(\angle B\).

8. Their intersection is \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

9. Draw perpendicular from \(I\) to a side.

10. Using that distance as radius draw the incircle.

✓ Incircle obtained.

---

• \(AB = 9\) cm
• \(\angle CAB = 115^\circ\)
• \(\angle ABC = 40^\circ\)

Locate incentre and draw incircle.

Construction Steps

1. Draw:
\[
AB = 9 \text{ cm}
\]

2. At point \(A\), construct:
\[
\angle CAB = 115^\circ
\]

3. At point \(B\), construct:
\[
\angle ABC = 40^\circ
\]

4. Let rays intersect at \(C\).

Triangle formed.

---

To Locate Incentre

5. Draw angle bisector of \(\angle A\).

6. Draw angle bisector of \(\angle B\).

7. Let them intersect at \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

8. Draw perpendicular from \(I\) to side \(AB\).

9. With centre \(I\) and that perpendicular distance as radius, draw the circle.

✓ Incircle constructed.

---

• \(AB = BC = 6\) cm
• \(\angle B = 80^\circ\)

Locate incentre and draw incircle.

Construction Steps

1. Draw:
\[
AB = 6 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 80^\circ
\]

3. On the ray mark:
\[
BC = 6 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle formed.

---

To Locate Incentre

5. Draw angle bisector of \(\angle A\).

6. Draw angle bisector of \(\angle B\).

7. Their intersection point is \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

8. Draw perpendicular from \(I\) to side \(AB\).

9. With centre \(I\) and radius equal to perpendicular distance, draw the circle.

✓ Incircle obtained.

---

Solution

Exterior angle property:

\[
\text{Exterior angle}
=
\text{sum of two interior opposite angles}
\]

✓ Answer:
\[
\boxed{(2)\ \text{Interior opposite angles}}
\]

---

✓ Answer:
\[
\boxed{(3)\ 105^\circ}
\]

---

Triangles formed:

• ΔAOB
• ΔBOC
• ΔCOD
• ΔDOA

Several congruent pairs are formed.

✓ Answer:
\[
\boxed{(1)\ 6}
\]

---

Using straight angle property:

\[
35^\circ + x^\circ + 60^\circ = 180^\circ
\]

\[
x = 180^\circ -95^\circ
\]

\[
x = 85^\circ
\]

✓ Answer:
\[
\boxed{(4)\ 85^\circ}
\]

---

Given:

• \(\angle C = \angle D\)
• \(\angle B = \angle E\)
• \(\angle A = \angle F\)

Correct correspondence:
\[
A \leftrightarrow F,\quad
B \leftrightarrow E,\quad
C \leftrightarrow D
\]

✓ Answer:
\[
\boxed{(4)\ \triangle ABC \cong \triangle FED}
\]

---

A rhombus with equal diagonals becomes a square.

✓ Answer:
\[
\boxed{(3)\ \text{Square}}
\]

---

Property:
\[
\angle AOB
=
\frac{1}{2}(\angle C + \angle D)
\]

✓ Answer:
\[
\boxed{\left(2\right)\ \frac12(\angle C+\angle D)}
\]

---

A parallelogram with one right angle is a rectangle.

✓ Answer:
\[
\boxed{(2)\ \text{rectangle}}
\]

---

Properties of parallelogram:

• Opposite sides are equal.
• Opposite angles are equal.
• Adjacent angles are supplementary.

✓ Correct statement:
\[
\boxed{(4)\ \text{Both pairs of opposite sides are equal}}
\]

---

• \(3x-40\)
• \(x+20\)
• \(2x-10\)

Find x.

Using angle sum property:

\[
(3x-40)+(x+20)+(2x-10)=180^\circ
\]

\[
6x-30=180
\]

\[
6x=210
\]

\[
x=35^\circ
\]

✓ Answer:
\[
\boxed{(2)\ 35^\circ}
\]

---

Find ∠ORS.

Equal chords subtend equal angles.

In isosceles triangle ORS:

\[
\angle ROS = 70^\circ
\]

Remaining angle sum:

\[
\angle ORS
=
\frac{180^\circ-70^\circ}{2}
\]

\[
=55^\circ
\]

✓ Answer:
\[
\boxed{(3)\ 55^\circ}
\]

---

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

Half chord:
\[
\sqrt{25^2-15^2}
=
\sqrt{625-225}
=
20
\]

Full chord:
\[
2\times20=40\text{ cm}
\]

✓ Answer:
\[
\boxed{(3)\ 40\text{ cm}}
\]

---

Find ∠AOB.

Angle at centre is twice angle at circumference.

\[
\angle AOB
=
2\times40^\circ
\]

\[
=80^\circ
\]

✓ Answer:
\[
\boxed{(1)\ 80^\circ}
\]

---

• \(\angle A = 4x\)
• \(\angle C = 2x\)

Opposite angles are supplementary.

\[
4x+2x=180^\circ
\]

\[
6x=180^\circ
\]

\[
x=30^\circ
\]

✓ Answer:
\[
\boxed{(1)\ 30^\circ}
\]

---

Given:

• \(CE=ED=8\) cm
• \(EB=4\) cm

Find radius.

Let radius be \(r\).

Distance from centre to chord:
\[
OE=r-4
\]

Using Pythagoras:

\[
(r-4)^2+8^2=r^2
\]

\[
r^2-8r+16+64=r^2
\]

\[
80=8r
\]

\[
r=10\text{ cm}
\]

✓ Answer:
\[
\boxed{(4)\ 10\text{ cm}}
\]

---

If ∠QRS = 100°, find ∠TVS.

Opposite angles in cyclic quadrilateral are supplementary.

\[
180^\circ-100^\circ=80^\circ
\]

✓ Answer:
\[
\boxed{(1)\ 80^\circ}
\]

---

Opposite angles are supplementary.

\[
180^\circ-75^\circ=105^\circ
\]

✓ Answer:
\[
\boxed{(2)\ 105^\circ}
\]

---

• \(\angle ADC = 80^\circ\)
• \(DC\) produced to \(E\)
• \(CF \parallel AB\)
• \(\angle ECF = 20^\circ\)

Find ∠BAD.

Using cyclic quadrilateral property and parallel lines:

\[
\angle BAD=120^\circ
\]

✓ Answer:
\[
\boxed{(3)\ 120^\circ}
\]

---

Given:

• \(AD=30\) cm
• \(AB=24\) cm

Find distance of AB from centre.

Radius:
\[
15\text{ cm}
\]

Half chord:
\[
12\text{ cm}
\]

Using Pythagoras:

::contentReference[oaicite:1]{index=1}

\[
d^2+12^2=15^2
\]

\[
d^2+144=225
\]

\[
d^2=81
\]

\[
d=9\text{ cm}
\]

✓ Answer:
\[
\boxed{(2)\ 9\text{ cm}}
\]

---

• \(OP=17\) cm
• \(PQ=30\) cm
• \(OS \perp PQ\)

Find \(RS\).

Half chord:
\[
PS=15\text{ cm}
\]

Using Pythagoras:

\[
OS^2+15^2=17^2
\]

\[
OS^2+225=289
\]

\[
OS^2=64
\]

\[
OS=8\text{ cm}
\]

Radius:
\[
OR=17\text{ cm}
\]

\[
RS=17-8=9\text{ cm}
\]

✓ Answer:
\[
\boxed{(4)\ 9\text{ cm}}
\]

---